Trigonometry (11th Edition) Clone

$\color{blue}{s=\dfrac{11\pi}{6}}$
Note that $\dfrac{\pi}{6}$ is a special angle and $\sin{(\dfrac{\pi}{6})}=\frac{1}{2}$. RECALL: An angle and its reference angle have either the same trigonometric values or they differ only in signs. Since $s$ must be in $\left[\dfrac{3\pi}{2}, 2\pi\right]$, then the angle must terminate in Quadrant IV. Note that the angle of $\dfrac{11\pi}{6}$ is in Quadrant IV and its reference angle is $\dfrac{\pi}{6}$. Sine is negative in Quadrant IV. Thus, $\sin{(\frac{11\pi}{6})} = -\frac{1}{2}$ Therefore, if $\sin{s} = -\frac{1}{2}$ and $s$ is in $\left[\dfrac{3\pi}{2}, 2\pi\right]$, then $\color{blue}{s=\dfrac{11\pi}{6}}$.